Week 1 Challenge
- Question no.1
- Ans - Given that there is 2 truss & to find out most efficiant gravity load carrying truss.
- Out of two trusses truss A is Pratt truss & Truss B is Howe truss.
- From this two types of truss both truss are efficient for carrying gravity load.
- Pratt truss have their vertical members in compression & diagonal members in tension.
- But in howe truss reverse is true as diagonal members in compression & vertical memebers in tension.
- As they are similar in structures, their uses are generally same.
- Depending on geometry & loading pratt truss have more unloaded membar.
- So both trusses are used as vertical load application such as gravity load carrying efficient loading.
- Question no. 2
- Ans - Given that figures of 2 beams & 2 frames with P loading in question.
- To find out bending moment diagram (BMD) & deflected shape of figures.
- Pdf attached below with answers of beam and frames.
- Here we take W loading means P loading given in question.
- We used fixed end moment method for bending moment diagram & stiffness matrix method for deflected shape.
- EI is constant , δ is deflection, K is stiffness.
- Question no. 5
- Ans - Pdf attached below.
- Here we take the bridge problem which is given that Coefficient of Thermal expansion of steel = 12x10-6/°C
- An initial total length (L) of 2000 m at -20 ºC.
- We have to find out calculate the length difference at 40 ºC and its total length (L).
- Length difference at 40 ºC = 1999.04 m.
- Total length L = 2000.96 m.
- Total length at free expansion L = 2001.44 m.
- Question no. 4
- Ans - Given that there is a floor load = 10 kn/m^2.
- To find out load on interior beam & exterior beam,interior girder & exterior girder.
- Here length of interior beam = 8 m.
- Number of floor = 2 no.
- So each load 2 x 10 x 8 = 160 kn/m.
- So load on interior beam is 160 kn/m.
- Here length of exterior beam = 8 m.
- Number of floor = 1 no.
- So that each load on exterior beam 1x 10 x 8 = 80 kn/m.
- So load on exterior beam is 80 kn/m.
- Here length of interior girder = 6 m.
- Number of floors = 6 no.
- So each load 6 x 10 x 6 = 360 kn/m
- So load on interior girder is 360 kn/m.
- Here length of exterior girder = 6 m.
- Number of floors = 3 no.
- So that each load on exterior girder 3 x 10 x 6 = 180 kn/m.
- So load on exterior girder is 180 kn/m.
- Question no. 3
- Ans - Given that 4 cross section with dimensions.
- To find out moment of inertia (MOI), weight, cross section area.
- Moment of inertia (MOI) = I mm^4.
- Weight = W Kg.
- Cross section area = A mm^2.
- For 1 channel section given (C).
- Flange width = 80 mm
- Flange depth = 10 mm.
- Web thickness = 8 mm.
- Web depth = 180 mm.
- Total cross section area (A) ∑a = 80 x 10 + 80 x 10 + 180 x 8 = 3040 mm^2.
- Total cross section area (A) = 3040 mm^2.
- Centroidal distance for axis zz.
- Rectangle 1 = area = 80 x 10 = 800 mm^2.
- Rectangle 2 = area = 80 x 10 = 800 mm^2.
- Rectangle 3 = area = 180 x 8 = 1440 mm^2.
- Centriodal distance from left side rectangle 1 & 2 = 40 mm.
- Centriodal distance from left side rectangle 3 = 4 mm.
- Volume of section = rectanangle 1 area x centriodal distance 1 = 800 x 40 = 32000 mm^3.
- Volume of section = rectanangle 2 area x centriodal distance 2 = 800 x 40 = 32000 mm^3.
- Volume of section = rectanangle 3 area x centriodal distance 3 = 1440 x 4 = 5760 mm^3.
- Total volume for zz axis (V) ∑v = 32000 + 32000 + 5760 = 69760 mm^3.
- Density of steel = 7850 kg/m^3. = 7.85 x 10^-6 kg/mm^3.
- Weigth (W) = volume x density = 69760 x 7.85 x 10^-6 = 0.5476 kg.
- Weight (W) = 0.5476 kg.
- Moment of inertia (I) about zz axis.
- To find out moment of inertia (I) = Izz = Izz1 + Izz2.
- Izz = [80 x 10^3/12 + 80 x 10 x 95^2] + [8 x 180^3/12 + 8 x 180 x 0^2]
- Izz = [6666.66 + 7220000 + 3888000] = 11114666.66 mm^4.
- Moment of inertia Izz = 1111.4666 x 10^4 mm^4.
- For a. plate girder section (I).
- Given that I section of plate girder with welded connection.
- Top & bottom flange plate 30 x 300 mm.
- Web plate 8 x 1000 mm.
- Flange width = 300 mm
- Flange depth = 30 mm.
- Web thickness = 8 mm.
- Web depth = 1000 mm.
- Total cross section area (A) ∑a = 30 x 300 + 30 x 300 + 8 x 1000 = 26000 mm^2.
- Total cross section area (A) = 26000 mm^2.
- Suppose we take length = 1000 mm. = 1 m.
- Volume of top plate = 30 x 300 x 1000 = 9 x 10^6 mm^3.
- Volume of bottom plate = 30 x 300 x 1000 = 9 x 10^6 mm^3.
- Volume of web plate = 8 x 1000 x 1000 = 8 x 10^6 mm^3.
- Total volume for (V) ∑v = 9 x 10^6 + 9 x 10^6 + 8 x 10^6 = 26 x 10^6 mm^3.
- Density of steel = 7850 kg/m^3. = 7.85 x 10^-6 kg/mm^3.
- Weigth (W) = volume x density = 26 x 10^6 x 7.85 x 10^-6 = 204 kg.
- Weight (W) = 204 kg.
- Moment of inertia (I) of plate girder.
- To find out moment of inertia (I) = I = I1 + I2 + I3.
- I = [300 x 30^3/12] + [300 x 30^3/12] + [8 x 1000^3/12].
- I = [675000 + 675000 + 666666.66] = 2016666.66 mm^4.
- Moment of inertia I = 201.6666 x 10^4 mm^4.
- For b. reinforced section with top & bottom plate (I).
- Given that I section with plate of top & bottom.
- Top & bottom flange plate 12 x 160 mm.
- Web section ISMB 300 from steel table.
- Flange width = 140 mm
- Flange depth = 12.4 mm.
- Web thickness = 7.5 mm.
- Web depth = 300 mm.
- Ixx = 8603.6 x 10^4 mm^4.
- Iyy = 453.9 x 10^4 mm^4.
- Cross section area = 5626 mm^2.
- Weight = 44.2 kg.
- Total cross section area (A) ∑a = 12 x 160 + 12 x 160 + 5626 = 9466 mm^2.
- Total cross section area (A) = 9466 mm^2.
- Suppose we take length = 1000 mm. = 1 m.
- Volume of top plate = 12 x 160 x 1000 = 192 x 10^4 mm^3.
- Volume of bottom plate = 12 x 160 x 1000 = 192 x 10^4 mm^3.
- Total volume for (V) ∑v = 192 x 10^4 + 192 x 10^4 = 384 x 10^4 mm^3.
- Density of steel = 7850 kg/m^3. = 7.85 x 10^-6 kg/mm^3.
- Weigth (W) = volume x density = 384 x 10^4 x 7.85 x 10^-6 = 30.14 kg.
- Total Weight (W) = 44.2 + 30.14 = 74.34 kg.
- Moment of inertia (I) of reinforced section with top & bottom plate.
- To find out moment of inertia (I) = I = Ixx + Iyy.
- Ixx = [8603.6 x 10^4] + 2 x [160 x 12^3/12 + 160 x 12 x (150 + 6)^2].
- Ixx = 17953.23 x 10^4 mm^4.
- Iyy = [453.9 x 10^4] + 2 x [12 x 160^3/12].
- Iyy = 1273 x 10^4 mm^4.
- Moment of inertia I = Ixx + Iyy
- I = 17953.23 x 10^4 + 1273 x 10^4.
- I = 19226.23 x 10^4 mm^4.
- For c. ISMB 500 & ISMC 300 cap channel section (I & C).
- Given that I section & channel section (C) cap at top.
- Top & bottom flange plate 12 x 160 mm.
- ISMB 500 from steel table.
- Flange width = 180 mm
- Flange depth = 17.2 mm.
- Web thickness = 10.2 mm.
- Web depth = 500 mm.
- Ixx = 45218.3 x 10^4 mm^4.
- Iyy = 1369.8 x 10^4 mm^4.
- Cross section area = 11074 mm^2.
- Weight = 86.9 kg.
- ISMC 300 from steel table.
- Flange width = 90 mm
- Flange depth = 13.6 mm.
- Web thickness = 7.6 mm.
- Web depth = 300 mm.
- Ixx = 6362.6 x 10^4 mm^4.
- Iyy = 310.8 x 10^4 mm^4.
- Cross section area = 4564 mm^2.
- Weight = 35.8 kg.
- Total weigth = 86.9 + 35.8 = 122.7 kg.
- Total cross section area (A) ∑a = 11074 + 4564 = 15638 mm^2.
- Total Moment of inertia I = I1 + I2.
- For ISMB 500
- I1 = Ixx + Iyy = 45218.3 x 10^4 + 1369.8 x 10^4 mm^4.
- I1 = 46588 x 10^4 mm^4.
- For ISMC 300
- I2 = Ixx + Iyy = 6362.6 x 10^4 + 310.8 x 10^4 mm^4.
- I2 = 6673.4 x 10^4 mm^4.
- Total Moment of inertia I = I1 + I2.
- Total Moment of inertia I = 46588 x 10^4 + 6673.4 x 10^4 mm^4.
- Total Moment of inertia I = 53261.4 x 10^4 mm^4.